Question: Show that the solution to the initial value problem $$x'=x^2+cos(t^2)$$ $$x(0)=0$$ is defined for $\lvert t \rvert\le\frac{1}{2}$ and satisfies there $\lvert x(t) \rvert\le1$.
My Attempt: Let $f(t,x)=x^2+cos(t^2)$, then $x'=f(x,t)$.
The Lipschitz condition: $$\lvert \frac{f(t,x_1)-f(t,x_2)}{x_1-x_2}\rvert=\lvert \frac{x_1^2-x_2^2}{x_1-x_2}\rvert=\lvert x_1+x_2\rvert$$
Because $x(0)=0$ and $x'>0$ for $\lvert t \rvert\le\frac{1}{2}$, $$x(t)>0$$ However, I could not prove the Lipschitz condition that $\lvert \frac{f(t,x_1)-f(t,x_2)}{x_1-x_2}\rvert$ is bounded. For example, suppose $x$ takes the limit of infinity, $$\lvert \frac{f(t,x_1)-f(t,x_2)}{x_1-x_2}\rvert=\lvert x_1+x_2\rvert=\infty$$
I do not know how to proceed from here. Thanks in advance!
$f(t,x)$ is locally Lipschitz in $x$, but not globally. This is enough to guarantee existence and uniqueness on some interval $[0,T)$ with $T>0$ (possibly $\infty$.)
On any interval where the solution is defined we have $$ x'\le x^2+1. $$ Integrate this inequality to obtain un upper bound on $x$.